Namely, how many times does 5/4 go into 95/4? well, let's just divide,
Answer:
(A) 33.9 (B) 22.6
Step-by-step explanation:
We first need to round to the nearest dollar for both prices. Lets examine both numbers.
$30.<u>1</u>2 $19.<u>9</u>3
We can round $30.12 to $30 since the number in the tenths place is 1 and we can round $19.93 to $20 since there is a nine in the tenths place.
Now we must find the tax for each number. We can do that by multiplying each number by 0.13 since that is 13% in decimal form. Lets do that now
Now we can find the final price for each number by adding the tax to the rounded price.
Hence, (A) 33.9, (B) 22.6
Answer:
-5/2+-2
Step-by-step explanation:
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
